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I don't understand this definition of the Sobolev space $H^{r,s}(E)$ for a vector bundle:

Def. The section $s:M \rightarrow E$ is contained in the Sobolev space $H^{r,s}(E)$ if for any bundle atlas with the property that on compact sets all coordinate changes and their derivatives are bounded, and for any bundle chart from that atlas, $\phi: E_U \rightarrow U \times R^n$, we have that $\phi \circ s_{|U}$ is contained in $H^{r,s}(U)$.

I understand that the requirement "on compact sets all coordinate changes and their derivatives are bounded" is for compatibility reasons but I don't know what does $H^{r,s}(U)$ mean.

Would the definition of $L^2$ section be the same except for the requirement on the derivatives to be bounded?


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$W^{p,q}(U)=H^{p,q}(U)=\{u\in\mathcal{D}'(U):D^\alpha f\in L^q(U),\forall|\alpha|\leqslant p\}$, if $p,q\in\mathbb{N}$. For general case, it's defined using Fourier transform, which is equivalent to the above one in integer case. If $q=2$, we often denote $W^{p,2}=H^p$. – user18537 Aug 19 '12 at 12:18
but $\phi o s_{|U}$ goes to $U\times R^n$ instead of $R$, how should I understand $H^{p,q}(U)$? just requiring that when composing with any chart the total map (going to $R^m\times R^n$) is in your definition of $H^{p,q}(U)$? thanks – inquisitor Aug 19 '12 at 14:29
Er, you can just view $\phi\circ s|U $ as a function from $U$ to $\mathbb{R}^n$, and it's in $W^{p,q}(U)$ if and only if every component is in $W^{p,q}(U)$. That is, we can say such a $\mathbb{R}^n$ valued function lies in $W^{p,q}(U;\mathbb{R}^n)$. – user18537 Aug 20 '12 at 2:02
@inquisitor where is this definition from? I'd like also to study those spaces on vector bundles.. – PtF Dec 11 '14 at 19:58
@PtF I think it was from Jost's book on differential geometry/Riemannian geometry – inquisitor Jan 3 '15 at 21:06

$H^{r,s}(U)$ is the usual Sobolev space on the flat domain $U\subset\mathbb{R}^m$.

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But $U$ is an open set in $M$ not in $R^m$ I guess the definition then is for $f o \psi^{-1}$ where $(\psi,U)$ is a chart of $M$ – inquisitor Aug 19 '12 at 14:35
Yes, it is all fine because of the invariant property of Sobolev spaces under coordinate transformations with bounded derivatives. – timur Aug 19 '12 at 14:44
So basically are the space of sections that in any coordinate chart they have coordinate functions from $R^m$ to $R^m \times R^n$ that belong to the usual $H^{p,q}$, right? – inquisitor Aug 19 '12 at 16:53
@inquisitor: if I understood you correctly, it should be $R^m$ to $R^n$. – timur Aug 20 '12 at 20:22
no I meant to $R^m \times R^n$ since the map the section composed with the trivialization of the vector bundle goes to that space, am I right? – inquisitor Aug 21 '12 at 18:23

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